At a baseball game Gina purchased 6 hotdogs and 3 nachos for $24. Frank purchased 8 hotdogs and 1 nacho for $23. How much is one hotdog and one nacho?

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At a baseball game Gina purchased 6 hotdogs and 3 nachos for $24. Frank purchased 8 hotdogs and 1 nacho for $23. How much is one hotdog and one nacho?

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Answer 1 · 2017-06-15T20:08:39

1 Hot dog = $2.50; 1 Nacho = $3.00

Explanation:

Let $x$ $x$ be the cost of 1 hotdog
let $y$ $y$ be the cost of 1 nacho

$6 x + 3 y = 24$ $6x + 3y = 24$ (Equation 1)

$8 x + 1 y = 23$ $8x + 1y = 23$ (Equation 2)

Divide equation 1 by 3
$2 x + y = 8$ $2x + y = 8$ (Equation 3)

Subtract Equation 3 from Equation 2
$8 x - 2 x + y - y = 23 - 8$ $8x - 2x +y - y = 23 - 8$

Simplifying
$6 x = 15$ $6x = 15$

1 Hot dog = $2.50

Substitute the value calculated above into Equation 1
$15 + 3 y = 24$ $15 + 3y = 24$
$3 y = 9$ $3y = 9$
1 Nacho = $3.00

Answer 2 · 2017-06-15T20:08:47

See a solution process below:

Explanation:

Let's call the price of a hotdog: $h$ $h$

Let's call the price of a nacho: $n$ $n$

We can then write:

$6 h + 3 n = $ 24$ $6h + 3n = $24$

$8 h + 1 n = $ 23$ $8h + 1n = $23$

Step 1) Solve the second equation for $n$ $n$ :

$- 8 h + 8 h + 1 n = - 8 h + $ 23$ $-color(red)(8h) + 8h + 1n = -color(red)(8h) + $23$

$0 + 1 n = - 8 h + $ 23$ $0 + 1n = -8h + $23$

$n = - 8 h + $ 23$ $n = -8h + $23$

Step 2) Substitute $(- 8 h + $ 23)$ $(-8h + $23)$ for $n$ $n$ in the first equation and solve for $h$ $h$ :

$6 h + 3 n = $ 24$ $6h + 3n = $24$ becomes:

$6 h + 3 (- 8 h + $ 23) = $ 24$ $6h + 3(-8h + $23) = $24$

$6 h + (3 \cdot - 8 h) + (3 \cdot $ 23) = $ 24$ $6h + (3 * -8h) + (3 * $23) = $24$

$6 h + (- 24 h) + $ 69 = $ 24$ $6h + (-24h) + $69 = $24$

$6 h - 24 h + $ 69 = $ 24$ $6h - 24h + $69 = $24$

$(6 - 24) h + $ 69 = $ 24$ $(6 - 24)h + $69 = $24$

$- 18 h + $ 69 = $ 24$ $-18h + $69 = $24$

$- 18 h + $ 69 - $ 69 = $ 24 - $ 69$ $-18h + $69 - color(red)($69) = $24 - color(red)($69)$

$- 18 h + 0 = $ - 45$ $-18h + 0 = $-45$

$- 18 h = $ - 45$ $-18h = $-45$

$\frac{- 18 h}{- 18} = \frac{$ - 45}{- 18}$ $(-18h)/color(red)(-18) = ($-45)/color(red)(-18)$

$(color(red)(cancel(color(black)(-18)))h)/cancel(color(red)(-18)) = $2.50$

$h = $2.50$

Step 3) Substitute $$2.50$ for $h$ in the solution to the second equation at the end of Step 1 and calculate $n$ :

$n = -8h + $23$ becomes:

$n = (-8 * $2.50) + $23$

$n = -$20.00 + $23.00$

$n = $3.00$

Hotdogs cost: $2.50

Nachos cost $3.00

Answer 3 · 2017-06-15T20:10:28

Answer: $$5.50$

Explanation:

Let $h$ be the cost of one hotdog and $n$ be the cost of one nacho.

We can set up a system of equations:
$6h+3n=24$
$8h+n=23$

We can solve for $h$ and $n$ by using elimination by multiplying the second equation by $3$ :
$24h+3n=69$

Now, we can subtract the first equation from the second equation to cancel out the $n$ term:
$24h+3n-(6h+3n)=69-24$
$24h-6h=45$
$18h=45$
$h=45/18=5/2=2.50$

Finally, we can find $n$ by substituting into either of the original equations and solving. We will use the second equation:
$8(5/2)+n=23$
$20+n=23$
$n=23-20$
$n=3$

Therefore the cost of one hotdog and one nacho is $h+n=2.5+3=5.5$ or $$5.50$

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At a baseball game Gina purchased 6 hotdogs and 3 nachos for $24. Frank purchased 8 hotdogs and 1 nacho for $23. How much is one hotdog and one nacho?

3 Answers

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